Separation of variables.
x(x-3)dy+ydx=0,
x(x-3)dy=-ydx,
dy/y=-dx/(x(x-3)).
If -1/(x(x-3))=A/x+B/(x-3), then:
-1=A(x-3)+Bx=Ax-3A+Bx, so:
-3A=-1, A=⅓.
And A+B=0, so B=-⅓.
dy/y=⅓(dx/x-dx/(x-3)).
Integrate both sides:
ln|y|=⅓(ln|x|-ln(x-3))+C=ln|∛(x/(x-3))+C.
Constant C can be incorporated into the log if C=ln(a).
So y=a ∛(x/(x-3)), or y³=bx/(x-3) where b=a³, a constant.